Question

函数y=f(x)由方程xe^y+y-1=0所确定,求二阶导数d^2/dx^2

Answer 1

方程两边对x求导

e^y+xe^y*y'+y'=0

所以y'=(-e^y)/(xe^y+1)=-1/[x+e^(-y)]

再次对方程两边的x求导

2e^y*y'+xe^y*(y')^2+xe^y*y''+y''=0

y''=-y'*(xy'+2)/[x+e^(-y)]

=(xy'+2)/[x+e^(-y)]^2

=[x+2e^(-y)]/[x+e^(-y)]

=(xe^y+2)/(xe^y+1)

=1+1/(xe^y+1)

=1+1/(2-y)

即d^2y/dx^2=1+1/(2-y)